What is the maximal curvature of the curve y = ln cos x?
Follow Math Help Forum on Facebook and Google+
Find an expression for the curvature using Then find the maximum.
problems. y' = -sinx/cosx or -tanx y'' = -secx^2
Put those in to get $\displaystyle \kappa=\frac{\sec^2 x}{(1+\tan^2 x)^{3/2}}$ Use $\displaystyle 1+\tan^2 x=\sec^2 x$ and $\displaystyle \kappa=\frac{\sec^2 x}{\sec^3 x}$. So $\displaystyle \kappa=\frac{1}{\sec x }=\cos x$.
View Tag Cloud